What is the sum of all values of x that satisfy the equation (x2 -5x+5)^(x2 +4x-60) =1? Hint: There are three cases to consider
I think I figured out 2 of the cases. The first one being x2 +4x-60=0 while x2 -5x+5=whatever. The second is x2 -5x+5=1 and x2 +4x-60=whatever. But I cant think of a third case
(-1)^ even power = 1
To find the third case, we can consider the case where (x^2 - 5x + 5) = -1 while (x^2 + 4x - 60) = whatever.
Let's solve these two cases one by one:
Case 1: x^2 + 4x - 60 = 0
To solve this quadratic equation, we can use factoring or the quadratic formula. Let's use factoring in this case:
(x + 10)(x - 6) = 0
Setting each factor equal to zero gives us two possible values for x:
x + 10 = 0 --> x = -10
x - 6 = 0 --> x = 6
So, in this case, the two values of x that satisfy the equation are x = -10 and x = 6.
Case 2: x^2 - 5x + 5 = 1
Rearranging the equation, we get:
x^2 - 5x + 4 = 0
Again, we can use factoring:
(x - 4)(x - 1) = 0
Setting each factor equal to zero gives us two possible values for x:
x - 4 = 0 --> x = 4
x - 1 = 0 --> x = 1
So, in this case, the two values of x that satisfy the equation are x = 4 and x = 1.
Case 3: (x^2 - 5x + 5) = -1
In this case, we'll solve for x^2 - 5x + 5 = -1:
x^2 - 5x + 5 + 1 = 0
x^2 - 5x + 6 = 0
Using factoring again, we have:
(x - 2)(x - 3) = 0
Setting each factor equal to zero gives us two possible values for x:
x - 2 = 0 --> x = 2
x - 3 = 0 --> x = 3
So, in this case, the two values of x that satisfy the equation are x = 2 and x = 3.
Now, to find the sum of all values of x that satisfy the equation, we add up all the values we obtained from each case:
-10 + 6 + 4 + 1 + 2 + 3 = 6
Therefore, the sum of all the values of x that satisfy the equation is 6.